Subject: Re: more elegant way of doing this? (very simple)
From: David Carlisle <davidc@xxxxxxxxx>
Date: Wed, 16 Aug 2006 17:53:03 +0100
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unless you really need to always use a node test
AContribution rather than name()='AContribution'
It's namespace aware and likely more efficient.
Are you using xsl 1 or 2?
xpath 1:
sum(Records/Record/*[self::AContribution|self::BContribution|self::CContribution][number()=number()])
xpath2 you can do the same, or a bit more simply:
sum(Records/Record/(AContribution|BContribution|CContribution)[number()=number()])
David
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